Question

Suppose Halley's comet orbits the sun every 79 years with an eccentricity of 0.87. What is it's aphelion distance in

au's

Answer 1

it's aphelion distance is 18.41 AU

Step-by-step explanation:

Given data:

eccentricity 0.87

P is period of orbiting = 79

from Kepler's third Law.

P^2 = d^3

where P is the period of orbiting in years, and

d it is orbit semi-major axis in Astronomical Units

1 AU is the average distance between earth and sun

so if P =79, we have

79^2 = d^3

6241 = d^3

d = (6241)^(1/3) = 18.41 AU